How to use definition of limit to compute the derivative of |x| Using definition of limit, I need to show
$$\lim_{\epsilon \to 0} \frac {|x + \epsilon| - |x|}{\epsilon} = \frac {x}{|x| }, x \neq 0$$
How should I proceed to get out of the absolute value signs?
 A: HINT:
Multiply and divide by $$|x+\epsilon|+|x|$$
SPOILER ALERT:

$$\begin{align}\lim_{\epsilon \to 0}\left(\frac{|x+\epsilon|-|x)}{\epsilon}\right)&=\lim_{\epsilon \to 0}\left(\left(\frac{|x+\epsilon|-|x|}{\epsilon}\right)\left(\frac{|x+\epsilon|+|x|}{|x+\epsilon|+|x|}\right)\right)\\\\&=\lim_{\epsilon \to 0}\left(\frac{2\epsilon \,x+\epsilon^2}{\epsilon(|x+\epsilon|+|x|)}\right)\\\\&=\frac{x}{|x|}\end{align}$$

A: The easiest way is to just compute the derivative separately for $x<0$ and $x>0$. For $x<0$,
$$\lim_{\varepsilon\to 0}\frac{|x+\varepsilon|-|x|}{\varepsilon}=\lim_{\varepsilon\to 0}\frac{-x-\varepsilon+x}{\varepsilon}=-1.$$
For $x>0$,
$$\lim_{\varepsilon\to 0}\frac{|x+\varepsilon|-|x|}{\varepsilon}=\lim_{\varepsilon\to 0}\frac{x+\varepsilon-x}{\varepsilon}=1.$$
A: $$|x|=\begin{cases}x &\text{if $x>0$}\\-x &\text{if $x<0$}\end{cases}$$
So you just need to split into 3 cases: Positive, Negative, and Zero.
For $x=0$, does $\displaystyle\lim_{x\to 0^+}\frac{x}{|x|}=\lim_{x\to0^-}\frac{x}{|x|}?$
I leave the rest to you, as it should be easy.
A: For $x\ne 0$ we have:
\begin{eqnarray}
\lim_{h\to 0}\frac{|x+h|-|x|}{h}&=&\lim_{h\to0}\frac{(|x+h|-|x|)(|x+h|+|x|)}{h(|x+h|+|x|)}=\lim_{h\to0}\frac{|x+h|^2-|x|^2}{h(|x+h|+|x|)}\\
&=&\lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h(|x+h|+|x|)}=\lim_{h\to0}\frac{2xh+h^2}{h(|x+h|+|x|)}\\
&=&\lim_{h\to0}\frac{h(2x+h)}{h(|x+h|+|x|)}=\lim_{h\to0}\frac{2x+h}{|x+h|+|x|}=\frac{2x}{2|x|}=\frac{x}{|x|}.
\end{eqnarray}
